CHEFIHG - Editorial

PROBLEM LINK:

DIFFICULTY:

Easy-Medium

PREREQUISITES:

Breadth First Search, Graphs

PROBLEM:

Given is a matrix of dimensions $N\times M$ in which some cells are marked with ".", some with "", and one with "C". The "." indicates that the cell can be visited, "" denotes restricted cell and "C" denotes the capital city. We have to give a sequence of moves of length at max $10^5$ in terms of "L" (left), "R" (right), "U" (up) and "D" (down) such that a robot starting from any "." cell in the in the grid and following the sequence of moves visits the "C" cell at least once.

EXPLANATION:

This problem doesn't require much thinking beyond the brute force way to do it. However, it is implementation heavy.

It is given that our matrix can at maximum be of $20 \times 20$ cells. This clearly hints towards the algorithm being close to brute force. Let us think about all the "." cells one by one. To begin with, let us look at the farthest "." cell from the capital city. How far can it be from the capital city? It is given that all the border cells are "*". This means that there can be a maximum of 400 - (20 - 20 -18 - 18) = 324 cells at maximum which are "." marked.

This tells us that the shortest path from any "." cell to capital city is at max 324 instructions (instructions being "L", "R", "U", "D") long. We first make a string of moves for the cell "." that is farthest from the capital city. For all other "." cells, we will have to execute the the same moves as well. After all the commands are executed, the farthest one has reached the capital city, and the others are in some position in the grid (most likely, not in the position they were).

Then we pick another one, and make it reach the capital city. The sequence of moves it takes for this is appended to the string we had from before, and all the remaining "." cells are again moved accordingly. We continue doing this for each of the "." cells till we have made each of them reach the capital city.

What is the length of the string at the end of all these operations? We have to move at most 324 cells. As we calculated before, the maximum length for one movement is 324 again. So, in total, 324*324 = 104976. This is slightly more than $10^5$. How do reduce this to below $10^5$? Actually, if we reason a bit more, we will see that the shortest path from any place in the grid to the capital city is 324 in theory only; it is actually much less. This is because 324 will only be if during a path from a cell to the capital, one has to go over all other cells. But why would one do that? We can move in all directions, so we can simply go from one row to its neighbouring ones or one column to its neighbouring ones. This tells us that the path will definitely be much shorter, definitely shorter than something like 305. And 305 * 324 = 98820. This is less than $10^5$. So this works.

But what is the time complexity? For each of the "." cells, we need to first do a BFS to find the shortest move sequence to the capital city. Then, for all the remaining "." cells, we need to make them follow the same move sequence. There can be $\mathcal{O}(NM)$ cells of type ".". BFS takes $\mathcal{O}(NM)$. But moving all other "." cells on the same move sequence takes $\mathcal{O}(N^2M^2)$; hence, this is the costliest operation per "." cell. So the total complexity is $\mathcal{O}(N^3M^3)$. This is sufficient given that $N = M = 20$ in the maximum case.

We had discovered it during the testing, we tried some test cases which can break some of the randomized solution. It was really hard to generate test cases for it. I would be interested if someone has a good test case to break such solutions.

It seems the test cases are modified after the contest. It no longer accepts random generated output.
The tester's solution needs to be updated to pass the new test cases.
My submission(Copied from the one who tried random approach): https://www.codechef.com/viewsolution/10900542

really shocked to see that first soln. why do you put such question when you are not deterministic about solution. I even tried that random thing during contests but it failed system tests so should
i blame my luck or your question making abilities.

If do start bfs from point C and build a string array s[i][j].s[i][j] is the path from point C to point (i,j).After bfs just go to every (i,j) which is '.' and reverse the string and replace 'U' with 'D' and 'D' with 'U' and R with L and L with R. Add all strings which contain path from (i,j) to C .
complexity is O(n^2).
if u manipulate string arrays while calculating bfs then complexity will be O(N).

Can we select one non-prohibitted area which is near to the Capital city and then move to the capital city. This requires only two characters in the command but solution can be correct.
Please correct me if I am wrong because this way the solution can be very very easy.

I have a doubt regarding this question. It was mentioned that we can choose any non prohibited cell as the starting point. To reach C any of the U,D,L,R should be non prohibited. So, if I just start from any non prohibited cell that lies either U,or D,or L, or R of 'C' and then just take it to C using single instruction, why am I getting WA? Pardon me if I understood the question wronmg.

@karanaggarwal I started from the Capital and went to each and every city and then reversed the string and replaced L with R, U with D and vice-versa. For all my custom test cases the code worked fine. But still its not getting accepted (WA). If you could help me figure my mistake out, I would be grateful. Below is the link to my code.

Random note regarding bounds on total sum of path lengths - in order to have cell with distance 300 from capital, you need to have a cell with distance <=1, a cell with distance <=2, a cell with distance <=3 and so on - all of them will be on the path from our "bad cell" to the capital. With this observation you can safely decrease bound to something like 53k.

BTW, just curious - what's the input file on which this algo produces longest possible string, and what's the length of that string? :)

Sir I have one doubt what if by applying the 324 step on the further most,what if any random block becomes the further most(robot moves to further most block),then it is not necessary it will work by this approach.